How to hide the request param in OpenApi? I would like to hide user_agent from OpenApi UI.
I have a simple app:
from typing import Optional
from fastapi import FastAPI, Header
app = FastAPI()
@app.get("/items/")
async def read_items(
user_agent: Optional[str] = Header(None),
size: Optional[int] = Body(None)):
return {"User-Agent": user_agent}
FastAPI in version 0.73.0 (related PR) supports this functionality natively: https://fastapi.tiangolo.com/tutorial/query-params-str-validations/#exclude-from-openapi.
from typing import Optional
from fastapi import Body, FastAPI, Header
app = FastAPI()
@app.get("/items/")
async def read_items(
user_agent: Optional[str] = Header(None, include_in_schema=False),
size: Optional[int] = Body(None)
):
return {"User-Agent": user_agent}
You can customize OpenAPI schema for hiding arbitrary parameters. The example below hidesparamB:
from fastapi import FastAPI
from fastapi.openapi.utils import get_openapi
app = FastAPI()
@app.get("/")
def get_items(paramA: int, paramB: int):
pass
def custom_openapi():
if app.openapi_schema:
return app.openapi_schema
openapi_schema = get_openapi(
title="Custom title",
version="2.5.0",
description="This is a very custom OpenAPI schema",
routes=app.routes,
)
# Remove paramB
params = openapi_schema["paths"]["/"]["get"]["parameters"]
params = [param for param in params if param["name"] != "paramB"]
openapi_schema["paths"]["/"]["get"]["parameters"] = params
app.openapi_schema = openapi_schema
return app.openapi_schema
app.openapi = custom_openapi